(x+y)^2-x-y
有條因式分解唔識做(x+y)^2-x-y,點計先岩?
回答 (5)
✔ 最佳答案
(x+y)2-x-y
=(x+y)2-(x+y)
=(x+y)[(x+y)-1]
=(x+y)(x+y-1)
應該係咁,
=(x+y)^2-x-y
=(x+y)^2-(x+y) 抽左個負出黎,變左有(x+y) 所以就有 得抽
=(x+y)((x+y)-1)
參考: me
(x+y)2-x-y
=(x+y)2-(x+y)
=(x+y)[(x+y)-1]
=(x+y)(x+y-1)
參考: me
(x+y)^2-x-y
=(x+y)^2-x-y
=x(x+y)
參考: me
(x+y)^2-x-y
=(x+y)^2-(x+y)
=(x+y)(x+y-1)
收錄日期: 2021-04-19 20:38:26
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